【python3】leetcode 674. Longest Continuous Increasing Subsequence（easy）

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674. Longest Continuous Increasing Subsequence（easy）

Given an unsorted array of integers, find the length of longest continuous increasing subsequence (subarray).

Example 1:

Input: [1,3,5,4,7]
Output: 3
Explanation: The longest continuous increasing subsequence is [1,3,5], its length is 3. 
Even though [1,3,5,7] is also an increasing subsequence, it's not a continuous one where 5 and 7 are separated by 4. 

Example 2:

Input: [2,2,2,2,2]
Output: 1
Explanation: The longest continuous increasing subsequence is [2], its length is 1. 

Note: Length of the array will not exceed 10,000.

涉及到连续子串问题

1 最耗时的解法O(n2) 可跳过看2

class Solution:
    def findLengthOfLCIS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if(len(nums) == 0): return 0
        maxlen = 1
        nowlen = 1
        for i in range(len(nums)):
            for j in range(i,len(nums)-1):
                if(nums[j+1] > nums[j]):
                    nowlen += 1
                    if(nowlen > maxlen):maxlen = nowlen
                else:break
            nowlen = 1
        return maxlen

Runtime: 468 ms, faster than 0.94% of Python3 o

2 优化：O(n)，start，end标记子串

思路：用两个字段标记连续子串的开始start和结束的后一位end，下一次查找start移动到end的位置，end++

class Solution:
    def findLengthOfLCIS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if(len(nums) == 0): return 0
        maxlen = 1
        start = 0; end = 1
        while(end < len(nums)):
            while((end < (len(nums))) and (nums[end] > nums[end-1])):end += 1
            nowlen = end - start  
            if(nowlen > maxlen):maxlen = nowlen
            start = end
            end += 1
            
        return maxlen

Runtime: 68 ms, faster than 17.88% of Python3